Arithmetic Sequences and Arithmetic Series - Basic Introduction - By The Organic Chemistry Tutor
Transcript
| 00:00 | in this video , we're going to focus mostly on | |
| 00:02 | everything . Tick sequences now to understand what an arithmetic | |
| 00:07 | sequences . It's helpful to distinguish it from a geometric | |
| 00:10 | sequence . So here's an example of number from six | |
| 00:15 | sequence . The numbers 37 11 15 , 19 23 | |
| 00:24 | and 27 represents an arithmetic sequence . This would be | |
| 00:31 | a geometric sequence . three six , 12 24 48 | |
| 00:39 | 96 1 92 . Do you see the difference between | |
| 00:45 | these two sequences ? And do you see any patterns | |
| 00:48 | within them and the arithmetic sequence on the left ? | |
| 00:53 | And notice that we have a common difference . This | |
| 00:55 | is the first term . This is the second term | |
| 00:58 | . This is the 3rd , 4th and 5th term | |
| 01:01 | to go from the first term to the second term | |
| 01:04 | . We need to add for to go from the | |
| 01:07 | second to the third term we need to add for | |
| 01:11 | . And that is known as the common difference in | |
| 01:19 | the geometric sequence . You don't have a common difference | |
| 01:21 | . Rather you have something that is called the common | |
| 01:24 | ratio . To go from the first term to the | |
| 01:27 | second term . You need to multiply by two to | |
| 01:31 | go from the second term to the third term . | |
| 01:33 | You need to multiply by two again . So that | |
| 01:36 | is the R . Value . That is the common | |
| 01:38 | ratio . So in an arithmetic sequence the pattern is | |
| 01:44 | based on addition and subtraction . In a geometric sequence | |
| 01:48 | , the pattern is based on multiplication and division . | |
| 01:53 | Now the next thing that when you talk about is | |
| 01:56 | the mean , how to calculate the arithmetic mean and | |
| 01:59 | the geometric mean . The arithmetic mean is basically the | |
| 02:03 | average of two numbers . It's A Plus B divided | |
| 02:07 | by two . So when taken an arithmetic mean of | |
| 02:12 | two numbers within an arithmetic sequence , let's say . | |
| 02:15 | If we were to take The mean of three and | |
| 02:18 | 11 we would get the middle number in that sequence | |
| 02:21 | . In this case we would get seven . So | |
| 02:24 | if you would add three plus 11 and divide by | |
| 02:28 | 23 plus 11 is 14 . 14 divided by two | |
| 02:31 | gives you seven . Now let's say if we wanted | |
| 02:34 | to find The emergency meeting between seven and 23 it's | |
| 02:40 | going to give us the middle number Of that sequence | |
| 02:43 | which is 15 . So if you would add up | |
| 02:46 | seven Plus 23 divided by two , seven plus 23 | |
| 02:51 | is 30 30 divided by two is 15 . So | |
| 02:55 | that's how you can calculate the arithmetic mean and that's | |
| 02:57 | how you can identify it within In November six sequence | |
| 03:04 | . The geometric mean is the square root Of eight | |
| 03:08 | times being . So let's say if we want to | |
| 03:11 | find the geometric mean between three and six , It's | |
| 03:15 | going to give us the middle number of the sequence | |
| 03:17 | . Which is I mean if we were to find | |
| 03:20 | the geometric mean between three and 12 We will get | |
| 03:23 | the middle number of that sequence which is six . | |
| 03:28 | So in this case a history be is 12 . | |
| 03:31 | Three times 12 is 36 . The square root of | |
| 03:34 | 36 is six . Now let's try another example . | |
| 03:39 | Let's find the geometric mean Between six and 96 . | |
| 03:44 | This should give us the Middle # 24 . Now | |
| 03:52 | we need to simplify this . Radical 96 is six | |
| 03:59 | times 16 . six times 6 is 36 . The | |
| 04:05 | square root of 36 is six . The square to | |
| 04:07 | 16 is four . So we have six times 4 | |
| 04:12 | Which is 24 . So as you can see the | |
| 04:15 | geometric mean of two numbers within the geometric sequence will | |
| 04:19 | give us the middle number in between those two numbers | |
| 04:21 | in that sequence . Now , let's clear away a | |
| 04:25 | few things . The formula that we need to find | |
| 04:32 | , the f term of an arithmetic sequence is a | |
| 04:36 | seven is equal to a someone Plus and -1 times | |
| 04:41 | the common difference d . In a geometric sequence it's | |
| 04:45 | a suburban is equal to a one times are Race | |
| 04:50 | to the N -1 . Now let's use that equation | |
| 04:59 | to get the fifth term in arithmetic sequence . So | |
| 05:04 | that's going to be a 75 . A sub one | |
| 05:08 | is the first term which is three , N is | |
| 05:13 | five . Since we're looking for the fifth term , | |
| 05:14 | the common difference is four . In this problem 5 | |
| 05:20 | -1 is four . four times 4 . 16 . | |
| 05:24 | 3 plus 16 is 19 . So this formula gives | |
| 05:31 | you any terms in the sequence . You could find | |
| 05:33 | the fifth term , the seventh term , the 100 | |
| 05:36 | term and so forth . Now in a geometric sequence | |
| 05:46 | , we could use this formula . So let's calculate | |
| 05:49 | the 6th term Of the geometric sequence . It's going | |
| 05:52 | to be a sub six Which equals a sub one | |
| 05:55 | . The first from history . The common ratio is | |
| 05:58 | too And this is gonna be raised to the 6 | |
| 06:02 | -1 . 6 -1 is five and then two to | |
| 06:07 | the fifth power . If you multiply 25 times two | |
| 06:10 | times two times two times two times two . So | |
| 06:15 | we can write it out . So this here that's | |
| 06:17 | for three twos . Make 84 times eight is 32 | |
| 06:22 | . So this is three times 32 . Three times | |
| 06:26 | 30 is 93 times two is 6 . So this | |
| 06:30 | will give you 96 . So that's how you can | |
| 06:35 | find the F term in a geometric sequence . By | |
| 06:41 | the way , make sure you have a sheet of | |
| 06:43 | paper to write down these formulas . So that when | |
| 06:46 | we work on some practice problems , you know what | |
| 06:49 | to do now the next thing we need to do | |
| 07:05 | is be able to calculate the partial sum of a | |
| 07:08 | sequence . S seven is the partial sum of a | |
| 07:15 | series of a few terms and it's equal to the | |
| 07:19 | first term plus the last term Divided by two times | |
| 07:24 | . And For geometric sequence the partial sum s event | |
| 07:28 | is going to be a sub one Times 1 - | |
| 07:32 | are basically and Over 1 -2 . So let's find | |
| 07:38 | the sum of the first seven terms in the sequence | |
| 07:43 | . So that's going to be S sub seven . | |
| 07:46 | That's going to equal the first turn plus the 7th | |
| 07:48 | turn divided by two times End . Where N . | |
| 07:54 | is the number of terms which is seven . Now | |
| 07:57 | think about what this means . So basically to find | |
| 08:00 | the sum of an arithmetic sequence , you're basically taking | |
| 08:04 | the average of the first and the last term in | |
| 08:08 | that sequence and then multiplying it by the number of | |
| 08:11 | terms in that sequence . Because this is basically the | |
| 08:15 | average of three and 27 . And we know the | |
| 08:21 | average or the arithmetic mean of three and 27 . | |
| 08:24 | That's gonna be the middle number 15 . So let's | |
| 08:27 | go ahead and plug this in . So this is | |
| 08:29 | 3-plus 27 . Over to times seven . Three plus | |
| 08:35 | 27 is 30 plus two . I mean well 30 | |
| 08:38 | divided by two . That's 15 . So the average | |
| 08:42 | Of the first and last term is 15 times seven | |
| 08:46 | . 10 times seven is 75 times seven is 35 | |
| 08:50 | . So this is gonna be one of five . | |
| 08:53 | So that's the some of the first seven terms . | |
| 09:00 | And you can confirm this with your calculator if you | |
| 09:02 | add up three plus seven plus 11 plus 15 Plus | |
| 09:08 | 19 plus 23 And then plus 27 . And that | |
| 09:13 | will give you s . f . seven . The | |
| 09:15 | some of the first seven terms . Go ahead and | |
| 09:22 | add up those numbers . If you do you'll get | |
| 09:25 | one of five . So that's how you can confirm | |
| 09:28 | your answer . Now let's do the same thing with | |
| 09:33 | a geometric sequence . So let's get this some Of | |
| 09:38 | the 1st 6 terms As sub six . So this | |
| 09:43 | is going to be three Plus six plus 12 plus | |
| 09:48 | 24 plus 48 plus 96 . So we're adding the | |
| 09:55 | first six terms now because it's not many terms were | |
| 10:04 | added we can just simply plug this into our calculator | |
| 10:08 | And we'll get 189 . But now let's confirm this | |
| 10:11 | answer using the formula . So as sub 6 to | |
| 10:16 | some of the first six terms Is equal to the | |
| 10:19 | first term . A sub one which history Times one | |
| 10:24 | -R . R . is the common ratio . Which | |
| 10:25 | is to race to the end and it's six over | |
| 10:30 | one minus R . Or one minus two . I'm | |
| 10:33 | gonna work over here since this more space Now 2 | |
| 10:37 | 6 , That's gonna be 64 . If you recall | |
| 10:41 | two to the fifth , power is 32 . If | |
| 10:43 | you multiply 32 x two you get 64 . So | |
| 10:48 | this is gonna be 1 -64 And 1 -2 is | |
| 10:52 | -1 . So this is three times 1 -64 Is | |
| 11:01 | -63 . So we could cancel the two negative science | |
| 11:05 | A negative divided by a negative will be a positive | |
| 11:08 | . So this is just three times 63 , three | |
| 11:10 | times six is 18 . So three times 60 has | |
| 11:14 | to be 1 80 and then three times three is | |
| 11:17 | 91 80 plus nine Adds up to 189 . So | |
| 11:23 | we get the same answer . Now , what is | |
| 11:27 | the difference between a sequence in a series ? I'm | |
| 11:32 | sure you heard of these two terms before . But | |
| 11:34 | what is the difference between them ? Now ? We've | |
| 11:37 | already considered what ? And the different sects sequences . | |
| 11:41 | A sequence is basically a list of numbers . So | |
| 11:47 | that's a sequence . A series is the sum of | |
| 11:51 | the numbers in the sequence . So this here is | |
| 12:01 | and arithmetic sequence . This is an arithmetic series because | |
| 12:06 | it's the sum of an arithmetic sequence . Now , | |
| 12:28 | what we have here is a sequence but it's a | |
| 12:31 | geometric sequence as we've considered earlier . This is a | |
| 12:37 | geometric series . It's the sum of a geometric sequence | |
| 12:42 | . Now there are two types of sequences and two | |
| 12:45 | types of series . You have a finite sequence and | |
| 12:49 | an infinite sequence and is also a finite series . | |
| 12:52 | In an infinite series . This sequence is finite , | |
| 12:59 | it has a beginning and it has an end . | |
| 13:01 | This series is also finite . It has a beginning | |
| 13:04 | and has an end . In contrast , if I | |
| 13:07 | would write 37 11 , 15 19 and then dot | |
| 13:12 | dot dot this would be an infinite sequence . The | |
| 13:20 | presence of these dots tells us that the numbers keep | |
| 13:23 | on going to infinity . Now the same is true | |
| 13:28 | for serious . Let's see if I had three plus | |
| 13:38 | seven Plus 11 plus 15 plus 19 and then plus | |
| 13:44 | dot dot dot dot dot . That would also be | |
| 13:48 | an infinite serious . So now , you know the | |
| 13:51 | difference between the finite series and an infinite series . | |
| 13:55 | Now , let's work on some practice problems , described | |
| 13:58 | the pattern of numbers shown below . Is it a | |
| 14:02 | sequence or serious ? Is it finite or infinite ? | |
| 14:06 | Is it arithmetic , geometric or neither ? So let's | |
| 14:11 | focus on if it's a sequence or series . 1st | |
| 14:15 | part eight . So we got the numbers 4 , | |
| 14:18 | 7 , 10 , 13 , 16 , 19 . | |
| 14:21 | We're not adding the numbers were simply making a list | |
| 14:24 | of it . So this is a sequence . The | |
| 14:30 | same is true . For part B will simply listing | |
| 14:33 | the numbers . So that's a sequence in part C | |
| 14:36 | we're adding a list of numbers . So since we | |
| 14:39 | have a some this is going to be a series | |
| 14:42 | D . Is also a series E . That's a | |
| 14:47 | sequence for F . We're adding numbers . So that's | |
| 14:51 | a series and the same is true Fiji . So | |
| 14:56 | hopefully this example helps you to see the difference between | |
| 14:58 | a sequence in the series . Now let's move on | |
| 15:01 | to the next topic . Is it finite or is | |
| 15:05 | it infinite to answer that ? All we need to | |
| 15:11 | do is identify if we have a list of dots | |
| 15:14 | at the end or not Here . This ends at | |
| 15:17 | 19 . So that's a finite sequence . The dots | |
| 15:23 | here tells us it's going to go forever . So | |
| 15:24 | this is an infinite sequence . This one we have | |
| 15:32 | the dot . So this is going to be an | |
| 15:33 | infinite series . This ends at 162 , so it's | |
| 15:40 | finite . So we have a finite series . This | |
| 15:44 | is gonna be an infinite sequence . Next we have | |
| 15:51 | an infinite series And the last one is a finite | |
| 15:56 | serious . Now let's determine if we're dealing with and | |
| 16:03 | a different take geometric or neither sequences series . So | |
| 16:09 | we're looking for a common difference or common ratio . | |
| 16:13 | So for a notice that we have a common difference | |
| 16:16 | of 34 plus three is 77 plus three is 10 | |
| 16:21 | . So because we have a common difference , this | |
| 16:24 | is going to be and arithmetic sequence for be going | |
| 16:32 | from the first number two . The second number we | |
| 16:34 | need to multiply by 24 times two is 88 times | |
| 16:38 | two is 16 . So we have a common ratio | |
| 16:42 | which makes this sequence geometric yeah for answer choice C | |
| 16:51 | . Going from 5 to 9 that's plus four And | |
| 16:54 | from 9 to 13 that's plus four . So we | |
| 16:57 | have a common difference . So this is going to | |
| 17:00 | be not a arithmetic sequence but and arithmetic series for | |
| 17:07 | answer choice D . Going from 2 to 6 were | |
| 17:10 | multiplied by three And then six times street is 18 | |
| 17:15 | . So that's a geometric the geometric series . Now | |
| 17:22 | for e Going from 50 to 46 that's a difference | |
| 17:26 | of negative four And 46 - 42 . That's a | |
| 17:30 | difference of -4 . So this is arithmetic for f | |
| 17:39 | . We have a common ratio of 43 times four | |
| 17:42 | is 12 . 12 times four is 48 . And | |
| 17:47 | if you're wondering how to calculate D . N . | |
| 17:49 | R . To calculate D take the second term subtracted | |
| 17:52 | by the first term 7 -4 Street . Or you | |
| 17:56 | can take the third term subtracted by the second . | |
| 17:59 | 10 -7 is 4th . In the case of f | |
| 18:02 | if you take 12 divided by three , you get | |
| 18:04 | four 48 divided by 12 to get four . So | |
| 18:08 | that's how you can calculate the common difference or the | |
| 18:10 | common ratio . It's by analyzing the second term with | |
| 18:14 | respect to the first one . So since we have | |
| 18:17 | a common ratio , this is gonna be geometric Fergie | |
| 18:25 | . If we subtract 18 by 12 we get a | |
| 18:27 | common difference of positive 6 20 for minus 18 Gives | |
| 18:32 | us the same common difference of six . So this | |
| 18:36 | is going to be arithmetic . So now let's put | |
| 18:40 | it all together , let's summarize the answers . So | |
| 18:44 | for part a what we have is a finite of | |
| 18:47 | reference six sequence . Part B . This is an | |
| 18:51 | infinite geometric sequence . See we have an infinite arithmetic | |
| 18:56 | series . D is a finite geometric series . E | |
| 19:02 | . Is an infinite referencing sequence . F is an | |
| 19:06 | infinite geometric series . G is a finite arithmetic series | |
| 19:13 | . So we have three columns of information with two | |
| 19:16 | different possible choices . Thus to to the third is | |
| 19:20 | eight , which means that we have eight different possible | |
| 19:22 | combinations . Right now , I have seven out of | |
| 19:26 | the eight different combinations . The last one is a | |
| 19:29 | finite geometric sequence , which I don't have listed here | |
| 19:34 | . So now , you know how to identify whether | |
| 19:37 | you have a sequence or series If it's arithmetic or | |
| 19:40 | geometric and if it's finite or infinite number , two | |
| 19:45 | Rights of first four terms of the sequence defined by | |
| 19:48 | the Formula A seven is equal to three and -7 | |
| 19:54 | . So the first thing we're going to do is | |
| 19:56 | find the first term . So we're gonna replace end | |
| 19:59 | with one . So it's gonna be 3 -7 which | |
| 20:04 | is negative for . And then we're going to repeat | |
| 20:06 | the process . We're going to find the second term | |
| 20:08 | a sub two . So it's three times 2 -7 | |
| 20:13 | Which is -1 . Next we'll find a sub three | |
| 20:20 | . three times 3 is 9 -7 . That's too | |
| 20:22 | . And then the fourth term eights up four , | |
| 20:26 | That's going to be 12 -7 which is five . | |
| 20:30 | So we have a first term of negative four then | |
| 20:33 | it's negative one 25 and then the sequence can continue | |
| 20:40 | . So the comment difference in this problem is positive | |
| 20:43 | three . going from negative 1-2 . If you add | |
| 20:46 | three you'll get to And then 2-plus 3 is five | |
| 20:52 | . But this is the answer for the problem . | |
| 20:54 | So this is those are the first four terms of | |
| 20:58 | the sequence number three . Right ? The next three | |
| 21:02 | terms of the following arithmetic sequence . In order to | |
| 21:08 | find the next re terms , we need to determine | |
| 21:11 | the common difference . A simple way to find the | |
| 21:15 | common difference is to subtract the second term by the | |
| 21:18 | first term , 20 to -15 is seven . Now | |
| 21:26 | , just to confirm , we need to make sure | |
| 21:27 | that the difference between the third and the second term | |
| 21:30 | is the same . 29 -22 . There's also seven | |
| 21:37 | . So we have a common difference of seven . | |
| 21:39 | So we could use that to find the next three | |
| 21:41 | terms . So 36 plus seven is 43 43-plus 7 | |
| 21:47 | is 50 50 plus seven is 57 . So these | |
| 21:51 | are the next three terms of the arithmetic sequence . | |
| 21:55 | Here's a similar problem but presented differently . Right ? | |
| 21:59 | The first five terms of an arithmetic sequence Given a | |
| 22:03 | one in D . So we know the first term | |
| 22:07 | is 29 and the common difference is negative for . | |
| 22:11 | So this is all we need to write the first | |
| 22:14 | five terms . If the common difference is negative for | |
| 22:17 | then the next term is gonna be 29 plus negative | |
| 22:19 | four , which is 25 25 Plus -4 or 25 | |
| 22:25 | -4 is 21 . 21 , -4 , 17 17 | |
| 22:30 | -4 is 13 . So that's all we need to | |
| 22:32 | do in order to write the first five terms of | |
| 22:36 | the romantic sequence . Given this information number five , | |
| 22:41 | right ? The first five terms of the sequence defined | |
| 22:44 | by the following recursive formulas . So let's start with | |
| 22:49 | the first one part A . So we're given the | |
| 22:53 | first term . What are the other terms when dealing | |
| 22:58 | with recursive formulas we need to realize is that you | |
| 23:01 | get the next term by plugging in the previous term | |
| 23:05 | . So let's say . And this too When N | |
| 23:08 | is too this is a sub two and that's going | |
| 23:12 | to equal a sub minus one to minus one is | |
| 23:15 | one . So this becomes a sub one plus four | |
| 23:20 | . So the second term is going to be the | |
| 23:22 | first term three plus four , which is seven . | |
| 23:29 | So we have three as the first term , seven | |
| 23:32 | as a second terms and that let's find the next | |
| 23:34 | one . So let's plug in three for end . | |
| 23:37 | So this becomes a century , The next one . | |
| 23:43 | This becomes a sub three modest one or a sub | |
| 23:46 | two plus four . So this is seven Plus four | |
| 23:53 | , which is 11 . At this point we can | |
| 23:56 | see that we have an arithmetic sequence with a common | |
| 23:58 | difference of four . So they get the next two | |
| 24:01 | terms we could just add for It's gonna be 15 | |
| 24:04 | and 19 . So that's it for part eight . | |
| 24:11 | So when dealing with recursive formulas , just remember you | |
| 24:14 | get your next term by using the previous term now | |
| 24:18 | for part B it's gonna be a little bit more | |
| 24:20 | work . So plugging in and equals two . We | |
| 24:25 | have the second term It's going to be three times | |
| 24:29 | the first term plus two . The first term is | |
| 24:33 | to so three times two is six plus two . | |
| 24:36 | That gives us eight . So now let's plug in | |
| 24:40 | an equal stream . When industry we have this equation | |
| 24:47 | , a submarine is equal to three times a sub | |
| 24:49 | 2-plus 2 . So we're going to take eight and | |
| 24:54 | plug it in here to get the third term . | |
| 24:57 | So it's three times eight plus two . three times | |
| 25:01 | 8 is 24 plus two . That's 26 . Now | |
| 25:08 | let's focus on the fourth term when N is four | |
| 25:10 | . So this is going to be a sub four | |
| 25:12 | is equal to three times a sub three plus two | |
| 25:17 | . So now we're gonna plug in 26 for a | |
| 25:19 | sub three . So it's three times 26 plus two | |
| 25:27 | . Three times 26 is 78 plus two , that's | |
| 25:30 | going to be 80 . Now let's focus on the | |
| 25:36 | 5th term . So a sub five is going to | |
| 25:40 | be three times a sub four plus two . So | |
| 25:45 | that's three times 80 plus two . Three times eight | |
| 25:49 | is 24 . So three times 80 is to 40 | |
| 25:52 | plus two . That's going to be 242 to the | |
| 25:59 | first five terms are too eight , 26 80 And | |
| 26:07 | to 42 . So this is neither and arithmetic sequence | |
| 26:12 | . Nor is it a geometric sequence , Number six | |
| 26:17 | . Right . A general formula or explicit formula , | |
| 26:21 | which is the same for the sequence is shown below | |
| 26:25 | in order to write a general formula or an explicit | |
| 26:27 | formula . All we need is the first term and | |
| 26:30 | the common difference if it's an arithmetic sequence which for | |
| 26:34 | part A it definitely is . So if we subtract | |
| 26:39 | 14 by eight we get six and if we subtract | |
| 26:42 | 20 by 14 we get six . So we can | |
| 26:44 | see that the common difference Is positive six and the | |
| 26:49 | first term is eight . So the general formula is | |
| 26:54 | a seven is equal to a someone Plus N -1 | |
| 26:58 | times deep . So all we need is the first | |
| 27:01 | term and the common difference and we can write a | |
| 27:05 | general formula or an explicit formula . The first term | |
| 27:10 | is eight D . Is six . Now what we're | |
| 27:14 | gonna do is we're going to distribute six to end | |
| 27:16 | -1 . So we have six times then which is | |
| 27:22 | six n . And then this will be -6 . | |
| 27:25 | Next we need to combine like terms , so eight | |
| 27:28 | plus negative six or eight minus six that's going to | |
| 27:31 | be positive too . So the general formula is six | |
| 27:36 | n plus two . So if we were to plug | |
| 27:42 | in one this will give us the first term 86 | |
| 27:46 | times one plus two is eight . If we were | |
| 27:49 | to plug in four it should give us the fourth | |
| 27:51 | term 26 . 6 times four is 24 Plus two | |
| 27:56 | , that's 26 . So now that we have the | |
| 28:00 | explicit formula for part A what about the sequence in | |
| 28:03 | part B . What should we do if we have | |
| 28:07 | fractions ? If you have a fraction like this or | |
| 28:13 | a sequence of fractions and you need to write an | |
| 28:16 | explicit formula , try to separate it into two different | |
| 28:20 | sequences . Notice that we have in arithmetic sequence . | |
| 28:24 | If we focus on the numerator , that sequences two | |
| 28:29 | , 3 , 4 , 5 and six . For | |
| 28:34 | the denominator we have the sequence 357911 . So for | |
| 28:40 | the sequence on top the first term is to and | |
| 28:43 | we can see that the common differences one , The | |
| 28:45 | numbers are increasing by one . So using the formula | |
| 28:49 | A seven is equal to a sub one plus n | |
| 28:52 | minus one times . D . We have that . | |
| 28:55 | A sub one is 2 & D is one . | |
| 29:00 | If you distribute one to end -1 you're just going | |
| 29:02 | to get N -1 . So we can combine two | |
| 29:06 | and negative one Which is positive one . So we | |
| 29:10 | get the formula and plus one . Yeah . And | |
| 29:15 | you can check it when you plug in 11 plus | |
| 29:17 | one is two . So the first term is to | |
| 29:21 | If you were to plug in five five plus one | |
| 29:24 | at six that will give you the fifth term which | |
| 29:27 | is six . Now let's focus on the sequence of | |
| 29:32 | the denominators . The first term mystery . The common | |
| 29:37 | difference we could see us too . 5 -3 is | |
| 29:40 | 2 7 -5 is too . So using this formal | |
| 29:44 | again we have a seven is equal to a sub | |
| 29:48 | one . A sub one history plus And -1 times | |
| 29:52 | d . d . s . two . So let's | |
| 29:56 | distribute to to end -1 . That's gonna be two | |
| 30:00 | N -2 . And then let's combine like terms 3 | |
| 30:05 | -2 is a positive one . So a seven is | |
| 30:10 | going to be two N plus one . So if | |
| 30:18 | we want to calculate the first term we plug in | |
| 30:21 | one for n two times one is two plus one | |
| 30:24 | . It gives us three If we want to calculate | |
| 30:27 | the 4th term and it's four . Two times four | |
| 30:30 | is eight plus one . It gives us not . | |
| 30:35 | So you always want to double check your work to | |
| 30:36 | make sure that you have the right formula . So | |
| 30:39 | now let's put it all together . So we're going | |
| 30:48 | to write a seven and we're going to write as | |
| 30:51 | a fraction . The sequence for the numerator is n | |
| 30:54 | plus one . The sequence for the denominator is two | |
| 30:58 | , n plus one . So this right here represents | |
| 31:08 | the sequence that corresponds to what we see in part | |
| 31:11 | B . And we can test it out . Let's | |
| 31:16 | calculate the value of the third term . So let's | |
| 31:20 | replace and with three it's going to be three plus | |
| 31:22 | one Over two times ST plus one . three plus | |
| 31:27 | 1 is four , two times three is six plus | |
| 31:30 | one at seven . So we have four of the | |
| 31:32 | seven . If we wish to calculate the fifth term | |
| 31:35 | , it's going to be five plus one over to | |
| 31:39 | Times five Plus 1 . Five plus one is 62 | |
| 31:42 | times five is 10 plus one . That's 11 . | |
| 31:48 | And so any time you have to write an explicit | |
| 31:50 | formula given a sequence of fractions , separate the numerator | |
| 31:55 | and there has not been into two different sequences . | |
| 31:58 | Hopefully they're both arithmetic . If it's geometric , you | |
| 32:01 | may have to look at another video that I'm going | |
| 32:03 | to make soon on geometric sequences but break it up | |
| 32:07 | into two separate sequences and then write the formulas that | |
| 32:09 | way and then put the two formulas in a fraction | |
| 32:12 | and that's how you can get the answer number seven | |
| 32:16 | , write a formula for the end of term of | |
| 32:18 | the arithmetic sequences shown below . Surviving . The formula | |
| 32:24 | for the f term is basically the same as writing | |
| 32:26 | a general formula for the sequence or an explicit formula | |
| 32:31 | . So we need to identify the first term which | |
| 32:33 | we could see us five . And the common difference | |
| 32:37 | 14 -5 is nine 203 -14 is nine as well | |
| 32:44 | . So once we have these two we can right | |
| 32:46 | the general formula . So let's replace the first term | |
| 32:53 | a someone with five And let's replace d . with | |
| 32:57 | nine . Now Let's distribute 9 to end -1 . | |
| 33:05 | So we're going to have nine N -9 . Next | |
| 33:08 | let's combine like terms . So it's going to be | |
| 33:13 | nine n . and then 5 -9 is -4 . | |
| 33:19 | So this is the formula for the f term of | |
| 33:22 | the sequence . Now let's do the same for part | |
| 33:33 | B . So the first term is 150 . The | |
| 33:38 | common difference is going to be 143 -150 which is | |
| 33:42 | -7 . To confirm that if you subtract 1 36 | |
| 33:46 | by 1 43 you also get negative seven . Now | |
| 33:53 | let's plug it into this formula to write the general | |
| 33:55 | equation . So a seven is going to be 150 | |
| 34:01 | plus And -1 times d . Which is -7 . | |
| 34:06 | So let's distribute negative 7 to end -1 . So | |
| 34:12 | it's gonna be 1 50 minus seven . End and | |
| 34:15 | then negative seven times negative one . That's going to | |
| 34:18 | be positive seven . So a sub N is going | |
| 34:21 | to be negative seven N Plus 1 57 . Or | |
| 34:27 | you could just write it as 1 57 minus seven | |
| 34:33 | N . So that is the formula for the term | |
| 34:38 | of the arithmetic sequence . Now let's move on the | |
| 34:45 | part beat Calculate the value of the 10th term of | |
| 34:49 | the sequence . So we're looking for a sub 10 | |
| 34:53 | . So let's plug intent into this equation . So | |
| 34:56 | it's gonna be nine Times 10 -4 . nine times | |
| 35:00 | 10 is 90 90 -4 , is 86 . So | |
| 35:05 | that is the 10th term of the sequence in part | |
| 35:09 | A for part B . The 10th term is going | |
| 35:12 | to be 157 -7 times 10 . seven times 10 | |
| 35:18 | is 70 1 , 57 minus 70 . It's gonna | |
| 35:23 | be 87 . Now let's move on the part C | |
| 35:31 | fined the sum of the 1st 10 terms . So | |
| 35:36 | in order to find the sum we need to use | |
| 35:39 | this formula ECE Ben is equal to the first term | |
| 35:45 | plus the last term divided by two times the number | |
| 35:48 | of terms . So if we want to find the | |
| 35:51 | some of the 1st 10 terms , we need a | |
| 35:53 | sub one which we know it's five A sub N | |
| 35:59 | and is 10 . So that's eight of 10 . | |
| 36:02 | The 10th term is 86 divided by two Times the | |
| 36:06 | number of terms which is 10 five plus 86 is | |
| 36:11 | 91 91 divided by two . gives us an average | |
| 36:15 | of 45.5 of the first and last number And then | |
| 36:19 | times 10 We'll get a total sum of 455 . | |
| 36:24 | So that is the sum of the 1st 10 terms | |
| 36:27 | of this sequence . Now , for part B we're | |
| 36:33 | gonna do the same thing , calculate acceptance . The | |
| 36:36 | first term a someone is 150 . The 10th term | |
| 36:41 | is 87 , divided by two times the number of | |
| 36:46 | terms which is 10 , 1 50 plus 87 . | |
| 36:51 | That's 2 37 Divided by two , that's 1 18.5 | |
| 36:57 | times 10 We get a sum of 1185 . So | |
| 37:05 | now you know how to calculate the value of the | |
| 37:07 | and turn and you also know how to find a | |
| 37:09 | some of a serious number . Eight Find the sum | |
| 37:15 | of the 1st 300 natural numbers . So how can | |
| 37:20 | we do this ? The best thing we can do | |
| 37:22 | right now is write a series . zero is not | |
| 37:26 | a natural number , but one is . So if | |
| 37:28 | we write a list one plus two plus three and | |
| 37:32 | this is going to keep on going All the way | |
| 37:35 | to 300 , So to find the sum of a | |
| 37:39 | partial series , we need to use this equation as | |
| 37:42 | seven is equal to a sub one Plus A 7/2 | |
| 37:47 | times end . Now let's write down what we know | |
| 37:52 | , We know that ace of one . The first | |
| 37:54 | term is one we know and is 300 . If | |
| 37:59 | this is the first term , this is the second | |
| 38:00 | term , this is the third term . This must | |
| 38:02 | be the 300 term . So we know it is | |
| 38:07 | 300 and A seven or a sub 300 is 300 | |
| 38:12 | . So we have everything that we need to calculate | |
| 38:14 | the sum of the 1st 300 terms . So it's | |
| 38:17 | a sub one which is one plus a seven which | |
| 38:20 | is 300 over to Times the number of terms , | |
| 38:24 | which is 300 . So it's going to be 301 | |
| 38:28 | divided by two times 300 And that's 40 5150 . | |
| 38:36 | So that's how we can calculate the sum of the | |
| 38:39 | 1st 300 natural numbers in this series . Number nine | |
| 38:46 | , Calculate the sum of all even numbers from 2 | |
| 38:49 | to 100 inclusive . So let's write a serious two | |
| 38:55 | is even three is odd , so the next even | |
| 38:57 | number is four And then six and then eight All | |
| 39:01 | the way to 100 . So we have the first | |
| 39:05 | term . The second term is for The third term | |
| 39:09 | is 60 , 100 is likely to be the 50 | |
| 39:13 | of turn but let's confirm it . So what we | |
| 39:16 | need to do is calculate end And make sure it's | |
| 39:18 | 50 and not 49 and 51 . So we're going | |
| 39:23 | to use this equation to calculate the value event . | |
| 39:30 | So a seven is 100 . Let's replace that with | |
| 39:34 | 100 . Ace of one is to the common difference | |
| 39:45 | We can see 4 -2 is 2 6 -4 is | |
| 39:49 | too . So the common difference is to in this | |
| 39:52 | example and our goal is to solve for end . | |
| 39:56 | So let's begin by subtracting both sides by two 100 | |
| 40:02 | -2 is 98 And this is going to equal two | |
| 40:06 | times and -1 . Next we're going to divide both | |
| 40:10 | sides by two 98 , divided by two is 49 | |
| 40:16 | . So we have 49 is equal to end -1 | |
| 40:19 | and then we're gonna add one to both sides . | |
| 40:23 | So n is 49 plus one which is 50 . | |
| 40:28 | So that means that 100 is indeed the 50th terms | |
| 40:32 | . So we know that end is 50 . So | |
| 40:36 | now we have everything that we need in order to | |
| 40:37 | calculate the sum of the 1st 50 terms . So | |
| 40:41 | let's begin by writing out the Formula 1st . So | |
| 40:48 | the some of the 1st 50 of terms is going | |
| 40:50 | to be the first term which is two plus a | |
| 40:53 | some 50 . The last term which is 100 divided | |
| 40:56 | by two times then which is 50 . So two | |
| 41:02 | plus 100 . That's 102 divided by two . That's | |
| 41:04 | 51 51 times 50 Is 2550 . So that is | |
| 41:12 | the sum Of all the even numbers from 2 to | |
| 41:15 | 100 inclusive . Try this one determine the sum of | |
| 41:19 | all odd integers from 20 to 76 . 20 is | |
| 41:24 | even but the next number 21 is odd And then | |
| 41:29 | 23 , 25 27 . All of that Are odd | |
| 41:34 | numbers up until 75 . So a someone is 21 | |
| 41:39 | in this problem , the last number a sub N | |
| 41:46 | is 75 . And we know the common differences too | |
| 41:52 | because the numbers are increasing right . What we need | |
| 41:56 | to calculate is the value event . Once we could | |
| 41:59 | find end then we could find a some from 21-75 | |
| 42:06 | . So what is the value then ? So we | |
| 42:09 | need to use the general formula for In a different | |
| 42:13 | six sequence . So a sub n is 75 . | |
| 42:17 | A sub one is 21 and the common difference is | |
| 42:21 | too . So let's attract both sides by 21 , | |
| 42:27 | 75 -21 . This is going to be 54 , | |
| 42:35 | Dividing both sides by two , 54 , divided by | |
| 42:41 | two is 27 . So we get 27 is N | |
| 42:44 | -1 and then we're going to add one to both | |
| 42:47 | sides . So n is 28 . So a sub | |
| 42:52 | 28 is 75 , 75 is the 28th term in | |
| 42:58 | the sequence . So now We need to find a | |
| 43:01 | some of the 1st 28 terms . It's going to | |
| 43:04 | be a someone the first term Plus the last term | |
| 43:08 | or the 28th term , which is 75 divided by | |
| 43:11 | two Times the number of terms , which is 28 | |
| 43:21 | , 21 plus 75 . That's 96 Divided by two | |
| 43:25 | , that's 48 . So 48 is the average of | |
| 43:28 | the first and the last term . So 48 times | |
| 43:31 | 28 , That's 1,344 . So that is the sum | |
| 43:38 | of the 1st 28 terms . |
Summarizer
DESCRIPTION:
This video provides a basic introduction into arithmetic sequences and series. It explains how to find the nth term of a sequence as well as how to find the sum of an arithmetic sequence. It also discusses how to distinguish a finite sequence from an infinite series. It also includes a few word problems.
OVERVIEW:
Arithmetic Sequences and Arithmetic Series - Basic Introduction is a free educational video by The Organic Chemistry Tutor.
This page not only allows students and teachers view Arithmetic Sequences and Arithmetic Series - Basic Introduction videos but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.
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